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(D) The equation of the parabola is $y^2 = 4x$. The tangent to this parabola is $y = mx + \frac{1}{m}$.The equation of the ellipse is $x^2 + 4y^2 = 8$

Vedclass · Accepted Answer

both $(A)$ and $(B)$

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(D) The equation of the parabola is $y^2 = 4x$. The tangent to this parabola is $y = mx + \frac{1}{m}$.The equation of the ellipse is $x^2 + 4y^2 = 8$,which can be written as $\frac{x^2}{8} + \frac{y^2}{2} = 1$. The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$.Here,$a^2 = 8$ and $b^2 = 2$. Substituting $c = \frac{1}{m}$,we get $\frac{1}{m^2} = 8m^2 + 2$.Multiplying by $m^2$,we get $1 = 8m^4 + 2m^2$,or $8m^4 + 2m^2 - 1 = 0$.Let $m^2 = t$. Then $8t^2 + 2t - 1 = 0$. Solving for $t$,we get $(4t - 1)(2t + 1) = 0$. Since $t = m^2 > 0$,we have $t = \frac{1}{4}$,so $m = \pm \frac{1}{2}$.For $m = \frac{1}{2}$,$c = \frac{1}{1/2} = 2$. The tangent is $y = \frac{1}{2}x + 2$,which simplifies to $x - 2y + 4 = 0$.For $m = -\frac{1}{2}$,$c = \frac{1}{-1/2} = -2$. The tangent is $y = -\frac{1}{2}x - 2$,which simplifies to $x + 2y + 4 = 0$.Thus,both $(A)$ and $(B)$ are correct.

The equations of the common tangents to the ellipse $x^2 + 4y^2 = 8$ and the parabola $y^2 = 4x$ are

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